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HW Chapter 10:
14, 20, 26, 44, 52, 64, 74, 92.
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Source Transformation
• Source transformation in frequency domain
involves transforming a voltage source in series
with an impedance to a current source in parallel
with an impedance.
• Or vice versa:
Vs  Z s I s
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
Is 
Vs
Zs
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Example:
using the source transformation method.
We transform the voltage source to a current source and
obtain the circuit.
Convert the current source to a voltage source
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Thevenin and Norton Equivalency
• Both Thevenin and Norton’s theorems are applied
to AC circuits the same way as DC.
• The only difference is the fact that the calculated
values will be complex.
VTh  Z N I N
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ZTh  Z N
Example
•
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Find the Thevenin and Norton equivalent circuits at terminals a-b for each of the
circuit.
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Op Amp AC Circuits
• As long as the op amp is working in the linear range,
frequency domain analysis can proceed just as it does
for other circuits.
• It is important to keep in mind the two qualities of an
ideal op amp:
– No current enters either input terminals.
– The voltage across its input terminals is zero with negative
feedback.
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Example
.
if v =3 cos 1000t V
s
frequency-domain equivalent
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Example contd…
Applying KCL at node 1,
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Example
For the differentiator shown in Fig. obtainVo/Vs. Find vo(t) when
vs (t) = Vm sin ωt and ω = 1/RC.
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Example
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Application
• The op amp circuit shown here is known as a
capacitance multiplier.
• It is used in integrated circuit technology to create
large capacitances out of smaller ones.
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Capacitor Multiplier
• The first op amp is acting as a voltage
follower, while the second one is an inverting
amplifier.
• At node 1:
Ii 
Vi  Vo
 jC Vi  Vo 
1/ jC
• Applying KCL at node 2 gives
Vo  
R2
Vi
R1
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Capacitor Multiplier II
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• Combining these two gives:
 R 
Ii
 j 1  2  C
Vi
R1 

• Which can be expressed as an input
impedance. V
1
Zi 
i
Ii

jCeq
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Capacitor Multiplier III
• Where:
 R 
Ceq  1  2  C
R1 

• By selecting the resistors, the circuit can
produce an effective capacitance between its
terminal and ground.
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Capacitor Multiplier IV
• It is important to understand a limitation of this
circuit.
• The larger the multiplication factor, the easier it is for
the inverting stage to go out of the linear range.
• Thus the larger the multiplier, the smaller the
allowable input voltage.
• A similar op amp circuit can simulate inductance,
eliminating the need to have physical inductors in an
IC.
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Oscillators
•
•
•
•
It is straightforward to imagine a DC voltage source.
One typically thinks of a battery.
But how to make an AC source?
Mains power comes to mind, but that is a single
frequency.
• This is where an oscillator comes into play.
• They are designed to generate an oscillating voltage
at a frequency that is often easily changed.
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Barkhausen Criteria
• In order for a sine wave oscillator to sustain
oscillations, it must meet the Barkhausen criteria:
1. The overall gain of the oscillator must be unity or
greater. Thus losses must be compensated for by an
amplifying device.
2. The overall phase shift (from the output and back to
the input) must be zero
• Three common types of sine wave oscillators are
phase-shift, twin T, and Wein-bridge oscillators.
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Wein-bridge
• Here we will only consider the
Wein-bridge oscillator
• It is widely used for generating
sinusoids in the frequency range
below 1 MHz.
• It consists of a RC op amp with
a few easily tunable components.
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Wein-bridge II
• The circuit is an amplifier in a non-inverting configuration.
• There are two feedback paths:
– The positive feedback path to the non-inverting input creates
oscillations
– The negative feedback path to the inverting input controls the gain.
• We can define the RC series and parallel combinations as Zs
and Zp.
Z s  R1 
j
C1
Zp 
R2
1  j R2C2
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Wein-bridge III
• The resulting feedback ratio is:
Zp
V2

Vo Z s  Z p
• Expanded out:
V2
 R2C1

Vo   R2C1  R1C1  R2C2   j  2 R1C1 R2C2  1


• To satisfy the second Barkhausen criterion, V2 must
be in phase with Vo.
• This means the ratio of V2/Vo must be real
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Wein-bridge IV
• This requires that the imaginary part be set to zero:
o2 R1C1 R2C2  1  0
• Or
o 
1
R1 R2C1C2
• In most practical cases, the resistors and capacitors
are set to the same values.
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Wein-bridge V
• In most cases the capacitors and resistors are made
equal.
• The resulting frequency is thus:
1
RC
1
fo 
2 RC
o 
• Under this condition, the ratio of V2/Vo is:
V2 1

Vo 3
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Wein-bridge VI
• Thus in order to satisfy the first Barkhausen
criteria, the amplifier must provide a gain of 3
or greater.
• Thus the feedback resistors must be:
R f  2 Rg
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